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CHEMICAL EQUILIBIURM

CHEMICAL EQUILIBRIUM

PART 1: EQUILIBRIUM LAWAND EQUILIBRIUM CONSTANT

Obiectives

. Explain, in tems of rates of fonyad and reverse reactions, what is meant by a reversible reaciion anddynamic equilibrium:

. Deduce expressions for equilibdum constanb in terms of concentration, lG, and partial pressures, lg;

. Calculate the values of equilibrium constanb in terms of concentrations or padial pressures from appmpriatedata;

. Calculate the quantities present at equilibrium, given appropriate data:

Texts and References

. A-Level Guides, Chemistry, J G R Bdggs, Longman, 3d ed, Chapter I

. A-Level Chemistry, E N Ramsden, Stantey Thomes (Publishers) Ltd, 2m ed, pg 19t206\* . Chemistry in Aciion, Michael Freemantle, Macmillan Educational Ltd, pg 250 - 2*,25&261

IRREVERSIBLE REACTION

. Many chemical reactions go to comoletion.

. This means that the reaction continues until one of the reactants is completelyused up in the reaction. When this happens the reaction stops -- lt has gone tocompletion.

. Such reactions are irrevercible reactions.

A+B)C+D

This can be representedgraphically:

Examples:

0) F'! t",r! .,"*; I +w lvrl t.ns r.rt

(t\ cacDg . z{tl a, taClz + (Az + q > o

c.9E{)oo

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CHEMICAL EQUILIBIUR'N

REVERSIBLE REACTION

. ManJ othe-r,reactions . lo ,ort go t a.,. p lctn,r , because theproducts of the reaction themserves reaa@ctanis. -

' such chemical reactions that take place in both directions are called reversiblereactions. This is indicated by the reversible sign ,' !; ".

&'*'d "r'AtB <-

C+ Dbr.lultl ,r |i

This can be representedgraphically:

After time t, forward and backward (reverse) reactions occur at the san€ ra,L

Both the concentrations ofthe reactants (A and B) and products (c and D) remain. arttdnt . There are still certain quantities of reactants remaining(i.e. concentrations of A and B * 0).

The system is said to reach

Examples ofreversible reactions:

Other examples:

OH2rar < ' :ut

@ rro3 F- ,Jo: r oz

co

o

o

Eo

oE(1,

o

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CHEMICAL EQWUAIURM

DYNAMIC EQUILIBRIUM

. In a reversible reaction both forward and backward reactions continue indefinitelyand the system is said to reach dynamic equilibium.

. In an equilibrium mixture, both the fonrard and backward

reactions are still taking place.

. System does not undergo any observable change.

i.e. Concentration of each species (equilibrium mixture)

remains constant.

I (*la o( &"oa rih i Rrtr e{ gr.g "o4rl

,.rrr

Dynamic equilibrium

o oao

o

ot oo o

t- . Equilibrium can only be achieved in a closed svstem in which there is no exchange

of matter or energy between the system and its surroundings.

EQUILIBRIUM LAW AND EQUILIBRIUM CONSTANT, K

o The behavior of a reversible reaction at equilibrium can be described in terms ofequilibrium law. This enables the calculation of a quantity known as theequilibrium constant, K.

, . K gives the equilibrium concentration of reactants and products.

. K can be expressed in terms of:(1) Concentrations;(2) Partial pressures of gases; and(3) Rate constants.

(1) K in terms of Concentrations

Consider an equilibrium system:

aA+bBscC+dD

A, B, C and D represent chemical substances,a, b, c and d represent stiochiometric coefiicients.

ay,'Lt"rlvtLa, r

fc l ' IsJ '' f r ] ' [Bl '

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I I = eqm conc in mol dm 3


The above iswhich relatesmixture.

a mathematical expression of the lad

the concentrations of reactants and procluctsIn an equilibrium

a

(2)

The expression shows that no matter what startingand D, when equilibrium is established, the above

aalarl

It is possible to determine the concentrationsexperiments.

concentrations are for A, B, Cratio of concentrations will be

of the substances through

+ dD (s)

Where P^ is the partial pressure ofA, etc.

(Unit of gas: atm orpa)

K" is called the .uiriLr:,r!, oe.r+.* for the reaction.K" is a constant atUnit of l(" is ( ,*t t--

K in terms of Paftial pressures

(Applicable to reactions involving gases.)

aA (S) + bB (S) s cc (s)

?' c.Kr,

oe 9b'A ' ,E

Unit of Ko . l "n + o, \ ' -^ 'o-t

Partial pressures of gases can be calculated usino

Pr = Total pressure of gasesp = Mole fraction of X = nx/nr

(x "P) '

(X Dp )

(x,p)^(X"t) t

K in Terms of Rate Consfants

K may be derived from kinetics if we know the rate equations for both forward andreverse reactions.tl

B5Ctb

t .e.

Pr= Xr. x P1

Hence, Ko can also beexpressed in terms of 1 and p:

(3)

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e.g. A +


. rf.orr ol 'S,parl .cac{ioy , 1. k1 [tl[o)

' fub + asvs< nocl.t:* , r" = kb fc]

' At 4clrb,t,a th7- .t* of 6,we'A alt Grqx 'l.acloat qQ .+al',

Yt = f ,

ua Itft s] , rr[.]

tc1/ 1g116e1) = Fr/Ku

t ( = k l /Kb

' Y.,;r',; :; t,

) '4 z o 'w '1 "nNote: This expression is only true if the chemical reaction proceed in a single step. Manyreactions do not proceed in one step but proceed through a complicated mechanism.

O Why does the expression : K = k/k applicable to only single step reactions?

POSITION OF EQUILIBRIUM

r+A \-- K+3( r at h Jh4.r t )

o lf conversion of P and Q into R and S is large, the p.o.m. lies to the nght:

f+ A P+S

l * i I b l^ ' . )

Change in conditions e.g. pressure, temperature etc will affect the p.o.m.

NOTES ON EQUILIBRIUM CONSTANTS KC AND KP

. Equilibrium law only applies to system at equilibrium.

. ( and [$ are only affected by_t"rr+".e_J_-The effect of temperature depends on the enthalpy change ofthe reaction, AH.

. Magnitude of G and Ko provides a useful indication of the e)dent of a chemicalreaction.

. Defines as the proportion of products to reactants in the equilibrium mixture.

p+n , . , tP)fs ' l| , \ , , r ,1 +S ' t= 6Li

. lf conversion of P and Q into R and S is sma//, the p.o.m. lies to the /eft

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CHEMICAL EQUILTBruRM

Reaction has a skong@position of equilibrium lies to the right.

composition will consist of largelyreactants; position of equilibrium lies to the left.

10-3<K<103Significant amou@

P.o.m. can be affected byca,rtt.7-t.'o,, cr. . However, K is only affected by temperatura

' K for a reaction indicates the extent of a reaction, but gives no information aboutthe rate of reaction. rt teils us how far but not how fast-the reaction g;;;. :-'

Examples:

(1) Give the ( and units for the following reactions:

A sealed flask contained 2 mores of iodine, 19 mores of hydrogen iodide and 4 mores ofhydrogen_at equiribrium. carcurate the equiiibriu; """.tr.i,

tt for the reaction. Give theunits for K.

J

(2)

Solution:b+ tt"r va.,,.c

^,,r. ol ,i,l ot .ri.

(*t7 o* €4,".

H2+

ol rt fl61g 51 1.

-'snr. r t -

.4L

t4Z'n

t (c lo m q,*tr

2HI

l,tr 7,K.' F;m

,HI,

t7

I

[ , r ] '

tiilil

(b) PCl3 + Cl, - pcts

[r.r:] [c1,]tpttcf

[-r r."'r1 [,,] r"- ff^r t*"t]

' L^a u'zJ

. r.t f,,".,rr"-{lLu-rJ .

--l;r,il L.c/i' ' ' l

= ;= t l * t t - , ) . ' )

l i l lca* J

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>t. : 45, rs

CHEUTCAL EQUIUBIURIt'

3(a) When a mixture of 50 g ethanoic acid and 20 g ethanol is allorcd to readt equilihium atconstant temp, it ls that 0.469 mol of ethancic acid a.s found to be gesent. Calorlate theequilibrium constant for the Baction at this temp€ratw€.

Solution

t;l tr$', to

Idt l f ur .J , f f 'o.r13

ar ,\0r., al . 0.033 - r.4h .0, rfrl

- tn ' * ) t<; l ,oo - >

-* - r ,Q

' t r@'a)v

CH3COTH (4 + CTH5OH (r) 5 cH3corqHs(4 + HzO(4

0o

oo

0',.*h o,tt. /v

cH3corcrHs(4 + HrO (4

o0

t! tr

lU. .l ',Et .l ctll &o. tr . fd ur.l

: o. {ot ,.,rllr

,o

ft. o,ers

o..fJt - I

U

O,8t3-, = o. .+t l t r E o, ' l+(A'r" \ , t - 'e 4 . tc. t&. , V f . t )

qPrtq 3 o' + r\lv o. rb+,/va.a-thh

,- fttr"o,"x3l[u'01[crrr a, n][6Hsc*r]

Io 3.+I I o.rr+l

IJILTJ: +.02 I

l+lt+]

(b) 3.fl) moles of ethanol and 1 .00 mde of ethanoic acid were mixed together at thesame temperaturc. Calculate the amount of ethyl ethanoate produced at equilibrium.

SolutionCH3CO2H (0 +CrHsOH (/) 5

: i. '.ol|r. ! r. oo 3, oo

t . o0 - r

t1.c-r) *

,

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lc t[cx1o1c,ns] f r l no1

fel, to, u] fa,$ ruJ

Fc ; . { .or :

lc l r^- l0.ortc + ra.of - o

, . . +.42 0<l \ v s.1o! fL

\,,.- *. 4 ..r *o..y ,..o g

t+l[+]

c{, (ad b


(4) Dissociation of NaOr0.05 mol N2o1 is introduced into a vessel of 1 dm3 at 25"c. The totar pressure atequilibrium is 142.5 kpa. Calculate l$ for the dissociation of N2Oa.

Solution:

To caVdh r,r0at gcst'.r ,

1V ' nr1_ o.os x c. . ) l r :41

1r tgr

= t..t rr x |0s Pa

NzOr(9) s 2NOdg)

Pi o+ ^LO4

i

l rv ? o. l h

I t , r . (o.r f , r . , r

Ne 01 l9) 1z

tri id l i nel I o.og

| ' t .4r I (Pr ' - P)

: l ,a3q t lo l -?

PLaoq s t\r.3.i -9) + e? . !4).! rP.

p -_ tqt , S - vt , i _ t t .g l<?a

:N02 f t )

o

. :P

)

P{,, ' ta t? 'b . 31.2 lPo

?|t . .a+ . t2r ,1 - t f , ( ' r ! j , ,

Kr'tD, tr7 2t

= t3 ' t+ -

l t . tk la I

,876 / ,

P.r, "{

103,3

Self-Directed Leaming:

(5) The equilibrium mixture in a flask was found to consist of o.s0 mole of so2, 0.12 moleOz and 5.0. Toles of SO3. All three substances were gaseo;s. The pressuie in thewas 8.0 x 105 Pa.(il Calculate the partial pressures of SO2, 02 and SO3 in the flask.(b) Write an expression for Ko for the equilib,rium:

'2SO, (g) + o, (g) S 2SOs (g)calculate the varue of G for the mixtuie in the flask. state the units.(c)

Ans: 5.85 x 1O'3 Pa'1

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PART 2: LE CHATELIER'S PRINCIPLE

Obiectives:

. State Le Chatelieis Principle and apply it to deduce qualitatively (from appropriate information) the effects ofchanges in temperature, concentration, pre€sure, on a system at equilibrium:

. Deduce wtrether cfranges in temperature, concenfation or pressure or the presence of a catalyst affect thevalue of the eouilibrium consiant for a readion:

LE CHATELIER'S PRINCIPLE

Summarizes the effect on the composition of an equilibrium mixture whenconditions are changed.

Staftr -lrat a'c|t a g.rrtn rv' p?ri hL'i/ m rr rubj.c+ad {o a chet'0( ih ,.4}i'{<r"'r

[i,e. clor{rr b so'ic rrft vnl Sdo yg r.rth 4t .t"d\'rt t f.t.tJqr , atc.) - +ra

Note: The system cannot completely cancel the change in the erternal factor, but it moves inthe direclion that will minimize the change.

FACTORS AFFECTING POSITION OF EQUILIBRIUM

1. lnfluence in Concentration

. When the concentration of the reactant is increased or decreased in anequilibrium system, reaction occurs so as to &rrrcr d 'tnd.a'€ d$au.lg

lL. co^ar.tro,{,oon ut 1i"c+ r^folqnC

Self-Directed Learning:

(6) The equilibrium constant K for the esterification reaction = 4.00. Calculate the massethanol that must be mixed with 2.0 mol of ethanoic acid to produce 1.5 mol of ethylethanoate at equilibrium.

cH3corH(4+c2H5oH(4 5 CH3COTCTH5(/)+ HzO(4

Ans: 121 .0 o

. To ' wt.rr *| t changz ^ n coq.'*V'5t"vti

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CHEMICAL EQUTUBruRM

2.

a

Example: Hr+lz

er.rrr *'olqrl.r Of gasgs.

Hence, increase in pressure favors the reaction that produces fewer morecures ofgases and vice versa.

Nz+3Hz 5 2NHJ

lnfluence in Pressure

when total pressure is changed, concentration of ail species in the container wirlchange to reduce the change in pressure.

lncrease in pressure (by decreasing the vorume) brings morecures crosertogether. To rerieve stress of additionar pressure, ieaction occurs to produce

,to 'rAue -+lr. irtru* i^ fA'] , e+l6\^ pctu,. sL,p6 1 a5.j!b!_ -r- grs Ar,a no4 H I

= fur&.I, .r oeht6 {arourrlH- + t -----------.-:4a!_znL

New eqm composition : t H ,1 D FA rf rnr*.,r i ft) Lt c x,,r<s

c.9Eo

o

By LCP, to reduce tfre Aecrto produce more H2.

ga$-a/d, rct'eAx^ la,or,'.t .=)

Art ! : r -=+ lHr

Neweqm composition: fx,-l * furl 4.,.ol. ., L!r1;nr*,K" : ,a.star+

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Example:

10


Total moles: 4

NOTE: Change in pressure has no effect on the p.o.m for gaseous readions if theno. of molecules in both sides of the equation is the same.

Example: Hu + lz

3. Influence in Temperature

5 2Hl

When an equilibrium s)Nstem is subjectedto an increase temperature,tnot;v

the

i:::;1,i,._'.t: , ' l r , i l ' ' , . i : " ; " ,15. . . , . , . , . : . , , . . . , , , ,

Etfect

Pressuret

9n<. *\t plt llt ( *r rye {on l^as + gr.,v*lrlr r cq^.fq^f +. a gartr h r

on {rrc dgh,| n& I lg r-cP , ep'l{{.u ,,lr 3h$t .t +" J5!L , ,.,hrch nA*ce 'l*

lrtc{ nvrrlcr cf trolc..to so ar to &.C!4,. lhr- atat p^..${n -

Jt E',.raaf €n.b- &.or,tl'.ila + 3H- :5 z r.rlle

( pu) ( p,J ( P*,)

New eqm composition i 4r, I ltx" I o'.^r. \ 4,,p, i" 14at'K": ,axa*.

PressureJ

Since the left side of the equation has 4 particles compared to 2 particles onthe right side, by LCP, eqm will shifts to the lu lt to increase the totalno. of molecules so as to increase the total oressure.

) ba<Y,,t l ea,&r lo,",t d ,

d:. r 3$ r- '-r,!--- L\tH A

New eqm crmposition i (*, a ltr, iv, crc.s ,j n\1r )c.re'eG: r.-rtor't

rcacuo.a AnA v{.. Vtrt(.

brn aht.

2NO2 AH>O (Foruard reaction is endothermic)(Backward reaction is exothermic)

Example: Nzor

t*

o l!{-

11 cd71'- + lA, o6o (l,o)r" rcX

Change Effect

Temperaturet 4

u", $,b*n (h*fr b $, .@ +- @,t4 noa g-dr4c tvt alr $,,y1)

ar.tro'f ,rsqtk th & s*q&oo 0+ rr4(ra$ ad t rno t a la -[le tf*tn

AYt.

AJCiJCI 11


:+ hrr.rad ,tn 1hsF rqtor.rA .

Nl.ll- :=r ,^101New eqm composition : [^r, o.] Aac,eot, E\t)] r^.,r r *

@il l iv ' (z.34.

TemperatureJ

By LCP, equitibriumGhifts toGel,y r-vr, equlllDnum snlTIS IO Ine l?+l tO prOdUCe mOre reaCtantSGin;the backward readion results in r;te;Gd;;;"rgi "nJi";"

to increase thetemperature.

$(r.I(n$ ra.nb,^ lceerr.l r 9" ..- :1p,

New eqm composition : f^lno.] i,.61ta,.t , fi.ro:f d"rar:..

fu o,l'lC =

Ifr;l- r,ilr de c+r

Note:a

a

f ncrease in temperature favors the ca [4ra1,q recto rDecrease in temperature favors the ..orr,r,.,r. F,

Lt4t t)

Position of equilibrium and K .rn"^"., .

o Reason for changes in K:

Recall K can be expressed as rate constants, K = k/k.Increase in temperature increases the rate constants and rates of both forwardreactions rhroush qi[fpq!€nqgl.[kr = Ae-Emr, k. = Ae€'m where Er + EJ

Influence of Catalyst

and backward

4.

a po ridt affect the position of equilibrium and K.

o Reason:catalyst lowers the activation energy of both forward and backward reactions::) iv\ c'\1 ..r r4+( 4 lr,{" .h,oO / uoff,,I,.'a .r"a^c,.r cir.ctlg

rate of reaction.

\ ^41?rrL 0cr,{r, \rt .,r<r,r}.dirrt 4 .pa'c o a+ o,*r.f lt r,rr

= ftrJ,L\ar-,Pt4'\'r {^a

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CHEMICAL EQUIUBIURIT

5. lnfluence of lnerl Gas (at constant temperature)

1. At constant volume

r Total pressure of the system increases due to increase in number of gaseousparticles.

Omlnal Svstem After additlon of lnert oasea (a I

@o@o@

@@

\9

00f-->@

oo@

a

oa @

@O

@

^o( , ,60

Equilibrium System: @ S O

. Concentrations or partial pressures of reactants and products remain ur'rlonttA

+ No alft,t en +lu prrtrora ee qrr'!b;ut' , aa| K.

2. At constant pressure

. To maintain a constant pressure, reaction volume must increase.

Orqinal Svstom After addition of inert qases fll I

mr--->o

ao

iooi" a

o

@

a@ o

. Partial pressure ofeach gases decreases.

. This effect is the same as if the volume increases is brought by ,. l.r^'.4

+'r. taaprl lftsaac (effect due to pressure).

. System reacts to increase partial pressures of reactants/products by favoring

oroduction of more molectrles

o Position of equilibrium will shift in the direction producing 'rc ,'lr.rc,rr..

^l( r l

I

@:

:@o

o

o

o

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CHEMICAL EQUIUBTURM

SUMMARY OF THE EFFECTSEQUILIBRIUM SYSTEM

Consider the reaction:

OF CHANGING CONDITIONS ON AN

aA + bB s cc +

Changes imposecl on thesystem at Equilibium

Equilibi um position shifts

Concentrations ofA and/or B increase

Concentrations ofC and/or D increase

Pressure increases lf (c+d) < (a+b): To right

lf (a+b) < (c+d): To tefl

Very little effect, if any, onreactions in liquid solution.

Temperature increases lf fonrvard reaction isexothermic: To left

If forward reaction isendothermic: To right

K decreased

K increased

o Rate of both forward andbackward reactionsincreases.

. Hence, equilibrium isachieved faster.

. K is also affected bytemperature.

Addition of catalyst Equilibrium is achieved fasterdue to lowering of E".

Presence of inert gas:

1. Constant volume

2. Constant totalpressure

No change

lf (ctd) < (a+b): To Ieft

lf (a+b) < (crd): To right

No change

2. Volume must increase:=) partial pressures of

reactantVproductsdecrease)

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CHEUICAL ECNI'UB'Uffi

(7) The equlllbrium mixture in a 2.oodms flask at soo oc uras found to consist of 0.6q) motof SO2, 0.200 mol of Oz and 4.80 mol of SOa.

(i) Calculate the equilibrium constant lG at S00 .C for the Eaction.2SOz(g)+Ods)s2SOg(g)

Solutlon:

(i)

etlt qf :

2sodg)

0. lar

PY r p lT,

F' r ,3 t+Jl rrr. l - l

+ Or(S)

a t00

Iq,r l r rot PA

(o. ao) (& ai)( +1t)z, oc (o. t ) t

g.+) b x rr t ta

.qhr f'lt tllna?)I ,oo l , , r l t

. tgl.>* sttl 0a

( t s+ . :* r rot) t

:50s f5)

4.tto

* .Fl,

= 6+o -l

l-'3

5 2sO.(S)

.+, 80

$+g

9!'rrt s|rt 4 4.

Pdttal R'Dr1r + 0. ,

,x?

r l0 r , l t + l t ld =

Of r1. : o,Lc or.o

.r t - .+.* , . g.4, I rcf i

(c.(ry)-(ryI'(+)

' 4'1b x,o-s qo-t ,

p',t

ET_

(ii)

= 6rr. et rro,)9 (o.rrtrrot)

How marry moles of 02 must be added into the flask at 50O "C in order toincrease the amount of SO3 to 5.0O moles?

lo\r.r { llrrr . el'

ro. ro. tltar.+) talA + 90)

A"JClJC1 15

CHEI(ICALEQUI'-'BIUR',

L**rohrt ,L

a r01q) r

dr. xfo rrl r llFGro

T = o'{t

. otor tr rt,r : -!]!,

.?2

Kct

e|lr} 5 9,

ot ? :Jor

,.tc. g- {9rr)

-5 {c 'l

{.,O. I

z

r{ l*.n +o'Lg

-9, Oo

go1

(€) '

(ry)"(+)ut9. l 13 4,15

3'-6

I ' o,t&t /

(iiD lf 2.00 mol of SO2 ard 1.00 mol of 02 are mixed and allou/ed to read at S(D "C.wttat must the total pressure be for 80 % conversion to SO3?

zl0r tl) + Qt3) i* 10" Q)

. 7,*n rl I r,or l,oo o {d.

, f r r f * , , . ( . - t .Lc "- - ( t+) a(o&)q . , , . to =o-2o :1.61

(aan l \ . o, t to +o,to. t .60

? , .20

^ tlr lTf t=T

( ! , t ) ( l ,$r)( tb+o?3)

" +.ol ( ,.t ?. I

-- { ro

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HOMOGENOUS EQUILIBRIA

Equilibrium reactions in which participating substances present in one phase only.

Examples: Reactions in liquid or gas phase.

CHgCOzH (/) + CTHsOH (/) s CHsCOzCzHs(D + HzO (/)

NzOa(g) s 2NO2(s)

HETEROGENOUS EQUILIBRIA

. Equilibrium reactions in wtrich participating substances present in different phase.

Example: Ca 4J : ts't -""'>

ao o (s) r co:. ( j)

fcaol p. . .h-

l f a (n31

5 ?,- . lgli. .^ I

Note: What is meant by the concentrations of pure solid/liquid?

. Concentration of a pure solid/liguid = *Cgry (a constant)

. Amount of solid/liquid may change, but concentrations of pure solids/liquids, and ofsolvents of dilute solutions are constant.

. lt is obviously sensible to combine any constants with K itself.

. Therefore, crncentrations of pure solid/liquid do not appear in the equilibrium constantexoression.

. The equilibrium constant expression for a heterogeneous reaction is written insuch a way that the concentrations of solids/ pure liquids are excluded in theexDression.

Examples: K. = [co, ] l ( t - - ?ro,or

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CHEMICAL EQUTUBURU

(8) A mixture of iron and steam _was allowed lo reach equilibrium at 600 .c. The equilibriumpressures of hydrogen and steam were 3.2 kpa and 2:4 kpa respectivery. c"i*i"-r.e (,.

3Fe (s) + 4HrO (g) S FqOa (s) + 4Hz (9)

Solutlon: 1 o \. ( r . r )(p= :_(P*"Jt I

=t tb ' l

(9) The solubility of the silver harides in water decreases in the order Agcr > AgBr > Agr.Because of this difi.*19 in solubility, bromide ions will displace ciloride iins frori'sotiosilver chloride. The equilibrium constant lC for this reaciion:

AgCt (s) + Br (aq) g AgBr (s) + Ct(aq)

is 360 at 2s8K. rf 0-100 mordmi Br (aq) is added to sorid Agcr, what wi[ be theequilibrium @ncentrations of Br(aq) and'Ci(aq).

Solutlon:

A6cl ls) . Bv- lq)

-

Ag Br b) + cr- tq,,,Itriltrt corc I a, log o

At A cq- ' , ' l !o- i "

{IAr ' o ' l - i

. l6b

16 -r lor . 1

, [ - - 36t ,

t z o,oe{\)

fu. . .+ . l * .F = o, oqq?a , ,o lJ\- l

r-^r^l 0I g/ - d orr" t v.t _ 5 sqq*L

2.?? r 1e.e ,.;g-3

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CHEMICAL EQUIUAIURM

PART3: HABERPROCESS

Obiectives

. Describe and explain lhe conditions used in lhe Hab€. process. as an example of the importance of anunderstanding of chemical equilibrium in the chemical industry.

THE HABER PROCESS

THE MODERN MANUFACTURING PROCESS

Many important industrial reactions are reversible reactions.Two important principles in choosing a process for industrial reactions are:(1) Efficient and (2) Economically viable

The conditions in an industrial reversible reaction aim to:o Produe maximum amount of product (high yield);o Using minimum amount of time;o At minimum cost.

The cost can be minimized by:o Using cheapest reagenfs (air, water);o Making readion as rapid as possible (catalyst);o Avoiding very high temperature/pressure, if possible.

TFri tzHaberstudiedtheN2(g}+3H,(g))2NH:(g)cqui l ibr iumint}reear lyE 19O0s and concluded that direct ammonia sjmthesis should be possible.

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Howevcr, it was not until l9l4 ttrat the cngineering problems and catal56tE question had been solved by Carl Bosch, and arnmonia production began justE in time for ttle atart of World War I. Ammonia is the starting matcrial to makcf nitric acid, a vital material in t}re manufacture of the e.folosives TNl and

Enitroglyccr in.Thereforc, th is ist l roughttobethef i rst l .argc.scaleuseofam synthetic chemical for military purposes.

IE Haber's contract with t}re manufacturer of ammonia called for him to receive 1

-L

A Fritz Haber (1868-1934) pfennig per kilograrn of ammonia, and he aoonE became not only famous but rictr-

-untom:natcly, hc joined the Gcrman

E Chemical Warfare Servicc at tlrc start of World War 1 and became its director

-r-- in 1916. The primary mission of tfie service was to develop gas warfare, and in

1915 he supervised the frrst use of Clz at tie battle of Ypres. Not only was this a tragedy of modernwarfare but also to Haber personally. His wife pteaded with him to stop his work in this area, and whenhe refused, she committed suicidc.

In 1918 he was awarded the Nobel Prize for the ammonia qmthesis, but ttte choice was criticized becauseof his role in chemica-l warfare.

After world war 1, Haber did some of his best work, continuing on in thermodynarnics (Bom-Haber99!el Jvas a product of tllis period). However, because he had a Jewish background, Haber left Gerrnanyin 1933, worked for a time in Englald, and died in Swi rland in 1934.

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THE HABER PROCESS

Nz(g) + 3Hz(g) I 2NH3(g) aH = 46 kJmol-l

OPTIMUM CONDITIONS FOR SYNTHESIS OF AMMONIA

Conditions for the synthesis of ammonia have to be carefully chosen if the process is tobe efficient and thus economically viable.

3 most important considerations are:(1) yield, (2) rate of reaction and (3) energy.

(1) Effect of Pressure: High Pressure

Out ReasonDesired ) lncrease rate of

reaction

D Pos'rtion of equilibriumshifted to the right(higher yield)

F An increase in pressure brings gasmolecules closer together. They collidemore frequently and hence reac{ion isfaster.

> According to LCP, high pressure increasesthe relative amounts of NH3 at equilibrium,since the foMard reac{ion results in adecrease in the no. of moles of gas (4 mol) 2 mol).

Undesired ) Not economicallyviable

) Cost of industrial plant and maintenancebecome more expensive, e.g. pipes have tobe tougher to withstand high pressure.

trr#--lqffidm&d

Nitrogen fromfractionaldistillation of liquidalr

t "r*"*" 1I from I

I synthesls gas

I

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(3)

. Compromise between high yield and low cost.

(21 EffectofTemperature: LowTemperafrtre

o Compromise between high yield and low rate.

Catalyst

Reason :,,., a:, ti-.ltr. . :..

Desired F Increase rate ofreaction

) Synthesis can becanied out at a lowertemperature.

> Catalyst lowers activation energy of reactionand hence increase the no. of effectivecollisions.

> High rate able to achieve at lowtemperature.

Conclusion

. According to the principles of kinetics and chemical equilibrium, the followingconditions would favor the manufacture of NH3:(1) High pressure(2) Moderotetemperoture(3) Use of catolyst

o Hence, optimum conditions chosen that compromises between kinetics andchemical equilibrium for Haber Process are:

(1) Pressure :200 - 300otm(2) Temperature : 450 - 500'C is used to obtqin reasonable rote of reocfion.(3) Catalyst : Finely divided Fe cotolyst. Efficiency is enhonced by cddifion

of smoll omount of promoters: AlaOr ond KzO.

YIELD OF AMMONIA

. Despite these optimum conditions, yield of NHr is only 10 - 15%.

Outeome.'. ',, ...:, ReaSOn t., . :.,:.rrijliii: .r..:

Desired F Position of equilibriumshifted to the right(higher yield)

> According to LCP, lower temperature tbvorsthe production of more heat to counteractthe low temperature. Hence, the exothermicreaction (fonnard reaction) is favored.

Undesired > Rate of reaction isvirtually zero

> At low temperature, no. of energeticcollisions which lead to chemical reactionare drasticallv reduced.

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. This yield can be increased by allowing liquid NH3 formed to be drawn off thesystem so that position of equilibrium is shifted to the right in favor of productionof more ammonia.

' unreacted nitrogen and hydrogen can be recycled. Ail reactant gases eventua[yconverted to NH3.

USES OF AMMONIA

.. - . Make feftilizefsi o.ro,i r $+r cr.r , aurrritr- r,l iaa+c. Makg nitfic acid: r, ,qr\! r{$! t a. y sxt c,, ryo1p (ilrl

, l.[rt. ,r"o. ca^ U .r,tl c ieF ,^'..'iqr rr,,6. Make synthetic fibres: ngn^ , .400^_

The Haber Process. for the production of ammonia from its elements can be represented by theequation: Nz(g) + 3Hdg) s 2NH3(g)

ln the process, a mixture of nitrogen and hydrogen is passed over a catalyst. The NH3 isremov_ed ftom.the products and unreacted gases passed through the rea&ion veir"r "g"in.1,",T

tiT:-j: ll": ?19o1 gas is removed frJm thereaction vesJet lwhere it accumuiate"s;,(axt) State the catalyst normally used in the Haber process

(ii) What are the sources of the nitrogen and hydrogen?(iiD Suggest where the argon comes ftom.(iv) \a/hy is the NH3 removed from the products before passing the unreacted nitrogen

and hydrogen over the catalyst again?

Self-Directed Learning

Exercise:

(b) Figure below shows the percentage of NH. in the equiribrium mixture under differentconditions.

Prgssuray'alm

z

(i)

(ii)

Vvhich temperature, 200 oC or S00 oC, produces a larger percentage of NH3 in theequilibrium mixture? state one disadvantage of using-a temperature wefi berow soo oc.\A/hich pressure, s0 or 1SO atm, produces

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larger pjrcent"ie of NH3 in fheequilibrium mixture?State one disadvantage of using a pressure greater than 200 atm.

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